[Maths Class Notes] on Ceva’s Theorem Pdf for Exam

The Ceva’s theorem is a theorem on the triangles that lie in the Euclidean plane geometry. It considers the ratio of the side lengths of the triangle that are divided by cevians.

 

Take an example of a triangle DEF, it’s lines DO, EO and FO be sketched over its vertices to a common point that is O, so that it could meet the opposite sides at G,H and I.

 

Ceva’s theorem is a theorem of affine geometry, in the context that it may be stated and proved without the use of the concepts of angles, areas, and lengths (except for the ratio of the lengths of two given line segments which are collinear). Therefore, it is true for triangles in any affine plane over any field. The Ceva’s theorem is helpful in proving the concurrence of cevians in the triangles and is commonly used in the Olympiad geometry. In this article, we will learn about Ceva’s Theorem and the converse of Ceva’s Theorem in detail.

Statement of Ceva’s Theorem

Consider a triangle ABC with a point P that lies inside the triangle. Consider the lines AP, BP, and CP to hit the BC, CA, and AB at the points D, E, F, D, E, and F, respectively.

According to Ceva’s theorem,

AF/FB . BD/DC . CE/EA = 1

The converse of Ceva’s theorem is true as well. if the points D, E and F lie on the sides BC, CA, and AB, respectively, in a way that,

AF/FB . BD/DC . CE/EA = 1

then the lines AD, BE and CF are concurrent at the point P.

Ceva’s Theorem Proof

Consider the same triangle.

Since the triangles △AFP and △FBP possess the same altitudes, you can write,

AF/FB = [AFB]/[FBP] = [AFC]/[FBC]

When you subtract the triangle area of the second quality from the area of the first equality, you get,

AF/FB = [APC]/[BPC]

Similarly, 

BD/DC = [APB]/[APC] and CE/EA = [BPC]/[APB]

When you multiply the previous three equations together, you would get,

AF/FB . BD/DC . CE/EA = APC/BPC . APB/APC . BPC/APB

Converse of Ceva’s Theorem

There are multiple proofs of the theorem that have been made. Two amongst the other are discussed below: 

The first one is shown just by using the basic properties of the areas of a triangle. However, it is not that easy because many cases must be considered, that depends on the  position that point O has.

Another proof is shown using vectors and barycentric coordinates and it seems more natural and dependent on the case. It works on any field in any affine plane. 

First Approach

The left-hand side’s sign is positive for two reasons. When O is inside the triangle, in this case,  all three ratios seem positive. When O is outside the triangle, in that case, one is positive and the remaining two seem negative.

If you want to check the magnitude, the area of the triangle with a given height is proportional to the base of the triangle. 

Consider the following triangle.

As you know that according to Ceva’s theorem,

AF/FB . BD/DC . CE/EA = 1

You have to prove that the Cevians AD, BE and CF are concurrent at a point.

Let AD and BE meet at an arbitrary point P. Let the third Cevian through the point P be CK. According to the previous proof, you have

(BD/CD) . (CE/AE) . (AK/BK) = 1

But you have assumed that 

(BD/CD) . (CE/AE) . (AF/BF) = 1

Hence, you can say that

(AK/BK) = (AF/BF)

Therefore, it is obvious that K = F and you can conclude that AD, BE and CF are concurrent. 

It was Giovanni Ceva, who published this theorem in his work, De lineis rectis in 1678. But there is evidence that it was proved much earlier by the eleventh-century king of Zaragoza named Yusuf Al-Mu’taman ibn Hud.

Second Approach

The three points D,E,F which are not collinear along with a point O, which are on the same plane, Gets three unique numbers with respect to barycentric coordinates of O. These unique numbers when added together equals 1. 

In Ceva’s theorem, the point O should not belong to any line that passes through two vertices of the triangle. This shows that those three unique numbers are not equal to 0. 

After rearranging the last equation by replacing X from F , the new equation will have a vector on its left-hand side, with the same direction as the line EF. On its right-hand side, it will have the same direction as that of line DE.

 

As D,E and F aren’t collinear, the directions of these lines will be different. It Shows the two members of the equation are equal to the zero vector and it also shows that the signed ratio of the length which are collinear line segments are on the left-hand side of the fraction. 

Solved Examples

1. Prove that if X, Y, and Z are the midpoints of the sides, then the three cevians are concurrent.

Solution: If you consider the same triangle,

then you can say that D, E, and F are the midpoints of their respective sides BC, AC, and AB. Therefore, you can say that AE = EC, CD = DB and BF = FA.

This satisfies the Ceva’s theorem that says that

AF/FB . BD/DC . CE/EA = 1

Hence, the intersection occurs at the centroid, which proves that the three cevians are concurrent.

2. Prove that the cevians perpendicular to the opposite sides are concurrent.

Solution: Consider D, E, and F to be the feet of the altitudes.

You must know that BFC∼△BDA.

Similarly,

△AEB∼△AFC, and △CDA∼△CEB.

Hence, you can say that

BF/BD = BC/BA , AE/AF =  AB/AC and CD/CE = CA/BC

When you multiply all these three equations, you would get,

BF/BD . AE/AF . CD/CE =  BC/BA . AB/AC . CA/BC

If you rearrange the left-hand side, you would get,

AF/FB . BD/DC . CE/EA = 1

Therefore, you can say that the three altitudes are coinciding at one single point, which is the orthocenter.